Andrzej Kaczyński
An Anticrack in Transversely Isotropic Space
AN ANTICRACK IN A TRANSVERSELY ISOTROPIC SPACE
Andrzej KACZYŃSKI
Faculty of Mathematics and Information Science, Warsaw University of Technology, ul. Koszykowa 75, 00-662 Warszawa, Poland
[email protected]
Abstract: An absolutely rigid inclusion (anticrack) embedded in an unbound transversely isotropic elastic solid with the axis of elastic
symmetry normal to the inclusion plane is considered. A general method of solving the anticrack problem is presented. Effective results
have been achieved by constructing the appropriate harmonic potentials. With the use of the Fourier transform technique, the governing
system of two-dimensional equations of Newtonian potential type for the stress jump functions on the opposite surfaces of the inclusion
is obtained. For illustration, a complete solution to the problem of a penny-shaped anticrack under perpendicular tension at infinity is given
and discussed from the point of view of material failure.
Key words: Three-Dimensional Elasticity, Transversely Isotropic Material, Plane Anticrack, Singular Integral Equations, Stress Singularity
1. INTRODUCTION
2. STATEMENT OF THE PROBLEM
The study of the redistribution of stresses due to the presence
of different kinds of defects in elastic bodies merit attention
of specialists from many branches, such as geomechanics, metallurgy, material science. From the viewpoint of inhomogeneities
in solids, cracks characterized by the displacement discontinuity
and flat rigid inclusions (called anticracks) with the traction discontinuity are the two dangerous extremes. Although the crack problems have been studied extensively over the past 50 years, research on the corresponding anticrack problems have been rather
limited and mainly concentrated on two-dimensional problems
related to rigid line inclusions (see the monographs by Berezhnitskii et al. (1983) and Ting (1996)). The corresponding threedimensional analysis of a penny-shaped anticrack (absolutely rigid
circular lamella) has been performed to a much lesser extent.
Detailed account of elastostatic problems involving planar rigid
inclusions embedded in homogeneous media is given in the studies by Kassir and Sih (1968), Selvadurai (1982), Silovanyuk
(1984, 2000), Podil’chuk (1997), Rahman (1999, 2002),
Chaudhuri (2003, 2012), Shodja and Ojaghnezhad (2007),
and in the monographs by Mura (1981), Panasyuk et al. (1986),
Khai (1993), Rogowski (2006), Kanaun and Levin (2008).
This paper is devoted to a three-dimensional static problem
in linear elasticity theory of an infinite transversely isotropic body
containing an arbitrarily shaped rigid inclusion lying in a plane
of isotropy and subjected to some external loads. In Section 2
the basic equations with the potential representations of their
general solution are reported. Besides, the anticrack problem
is formulated. Section 3 presents a general method of solving
the resulting boundary value problems. The governing 2-D singular integral equations are obtained in terms of the stress discontinuities across inclusion faces. As an illustration, a closed-form
solution is given and discussed in Section 4 for a circular rigid
inclusion subjected to tension at infinity. Finally, Section 5 concludes the paper.
2.1. Basic equations and general potential solutions
140
The results of comprehensive research on the theory and applications of the mechanics of transversely isotropic elastic materials can be found in the book by Ding et al. (2006).
Referring to the rectangular Cartesian coordinate system
denote at the point
the displacement vector by
and the stresses by
,
,
,
,
,
.
Throughout the considerations the following notations will be
used: Latin subscripts always assume values 1, 2, 3 and the
Greek ones 1, 2. The Einstein summation convention holds and
subscripts preceded by a comma indicate partial differentiation
with respect to the corresponding coordinates.
Consider a homogeneous transversely isotropic space assuming that the axis of elastic symmetry of the material coincides with
the -axis. Then the number of independent elastic stiffness
constants (moduli) in the generalized Hooke’s low reduces
to five: , , ,
,
. The stress-displacement equations
are the following:
σα 3  c 44 uα ,3  u3,α  , α  1, 2
(1)
σ33  c13uγ ,γ  c33u33
,
(2)
σ11  c11 u11
,  c12 u22
,  c13 u33
,
(3)
σ 22  c12u11
,  c11 u22
,  c13 u33
,
(4)
σ 12  0,5 c11  c 12  u1,2  u21
, 
(5)
Inserting the above expressions into equilibrium equations in
the absence of body forces
σ jk ,k  0
(6)
acta mechanica et automatica, vol.7 no.3 (2013), DOI 10.2478/ama-2013-0024
a governing system of three linear partial differential equations
of second order with constant coefficients for displacements
is obtained:
0,5c 11  c 12 uγ ,γα  0,5 c 11  c 12  uα ,γγ  c 44 uα ,33 
 c 13  c 44  u33
, α  0 , α  1, 2
c13  c 44  uγ ,γ 3  c 44u3,γγ  c33u333
, 0
(7)
(8)
According to the results obtained by Kaczyński (1993) with
some modifications, the general solution of the above equations
is found in terms of three potential functions
,
with
which are harmonic in the appropriate
coordinate systems
, i.e.
2
φ j ,γγ 
 φj
z 2j
 0 , j  1,2,3; no sum on j
(9)
The form of displacement representation is dependent on the
three material characterizing parameters . The constant
is given as:
t3 
c11  c12  / 2c 44  0
For the sake of simplicity, hereinafter we proceed to consider
the case with distinct eigenvalues (Case 1). Then the displacements and stresses can be represented in terms of potentials as
follows:
uα  φ1  φ2 ,α  γα φ3 , γ , α  1, 2
u3  mγ t γ
φγ
(18)
zγ
2φγ
σ 3α
2φ3
 1  mγ t γ
 γα t 3
c 44
zγ x α
z 3 x γ
(19)
2φγ
σ 33
 1  mγ
c 44
zγ2
(20)






σ 11  c 44 1  mγ t γ2
2
c 33 c 44 t 4  c 13   2c 13 c 44  c 11c 33 t 2  c 11 c 44  0


Case 1: t 1  t 2 when A  2c 44

t 1  0,5t   t    0
A  2c 44  

t 2  0,5t   t    0

(12)

σ 22  c 44 1  mγ t γ2
zγ2

(21)
2φγ
zγ2

(22)
 c 11  c 12  φ1  φ2 , 11  φ3 ,12 


(11)
The analysis of modulus restrictions based on the positive
definiteness of the stress energy leads to two cases (Ding et al.;
2006):
2φγ
 c 11  c 12  φ1  φ2 , 22  φ3 ,12 


(10)
while the constants
are the roots with positive real
part of the following eigen-equation:
(17)


σ 12  c 11  c 12  φ1  φ2 , 12  0,5 φ3 ,11  φ3 ,22 


(23)
Here 12  1, 12  1 , 11 22  0 and the constants
related to
are defined as:
mα 
c 11  t α2 c 44
2
c 13  c 44  t α

c 13  c 44
2
t α c 33  c 44
, α  1, 2
(24)
It is worth noting that:
or

t 1  0,5t   i t *  2
A  2c 44  
, i  1

t 2  0,5t   i t * 
(13)
where:
t 1t 2  c 11 / c 33
A  c 11c 33  c 13
t 
 A  2c 44  A / c 33c 44
t* 
2c 44  A  A / c 33c 44
(14)
A  2c 44  t 1  t 2  4 c11 / c33  t 0
(15)
In the case when the material is isotropic it is found that:
c 11  c 33  λ  2μ 
c 12  c 13  λ 
Here,
2μ 1 ν 
(16)
1  2ν
2μν
, c 44  μ
1  2ν
are Lamé constants, and

(25)

2
 2c 13c 44 / c 33 c 44
t 1 2  t 2 2  c 11c 33  c 13
2.2. Formulation
Case 2: t 1  t 2
t0 t3  1
c m t 2  c  c 1  m 
α
 33 α α 13 44
, α  1 or 2, no sum on α

2
2
c 11  c 13 mα t α  c 44 t α 1  mα 
m1m2  1
is Poisson’s ratio.
Consider a transversely isotropic space with the axis of symmetry as the - axis and the isotropic plane as the
- plane.
Suppose that this body is weakened by a rigid-sheet like inclusion
(anticrack) occupying a certain domain with a smooth boundary
at the plane
and subjected to some external loads.
It is known that an anticrack problem can be regarded as the
superposition of two problems. One (labelled by the subscript 0)
is no inclusion problem with the given applied loadings and the
other is the perturbed problem in which the displacements along
the anticrack are prescribed as the negative of those generated
in the first problem. Thus, the total displacement-stress field (denoted by
and
) can be expressed as:
141
Andrzej Kaczyński
An Anticrack in Transversely Isotropic Space
t   u 0  u , σ t   σ 0  σ
j
jk
j
jk
jk
(26)
uj
 σ 3α  x 1 , x 2,0

S
  σ3α  x1,x 2,0  dx1dx 2  0
  σ31  x 1,x 2,0  dx1dx 2 
S
  x 1 σ 32  x 1 , x 2 ,0   σ 32  x 1 , x 2 ,0   dx 1dx 2  0


 x 2 σ 31  x 1 , x 2,0
(39)
Moreover, we assume that
and
)are known from the
solution to the 0-problem. As a matter of fact, only the values
of displacements
̅
are
needed in the subsequent analysis.
Next, we concentrate attention on the non-trivial perturbed
problem, solution to which tends to zero at infinity and satisfies
the necessary boundary conditions on :
In addition to these boundary conditions, the displacement
and stress fields must vanish at infinity.
0
u1  u1   ε1  ω3 x 2
(27)
3. SOLUTION METHOD
0
u2  u2   ε 2  ω3 x 1
(28)
0
u3  u3   ε3  ω2 x 1  ω1x 2
(29)
in which also the unknown small displacements and rotations
are included to describe a motion of the inclusion as a rigid
unit. These parameters will be determined later in solving the
problem in hand from the equilibrium conditions of the anticrack
(no resultant forces and zero-moments).
Now we proceed to the reduction of the above anticrack
problem to some mixed boundary-value problems of potential
theory related to the one of a half-space (say,
). From the
relevant symmetry properties about the plane
follows the
division of the problem into antisymmetric (A) and symmetric ones
(B).
In the antisymmetric part (A) of the problem, the mixed
conditions on the plane of antisymmetry may be written as

(40)
S
The above-mentioned problems (A) and (B) are now reducing
to some mixed boundary-value problems of potential theory
by constructing the potentials functions well suited to the boundary conditions (30)-(32) and (35)-(38), respectively. Further,
an integral equation formulation is given for these problems.
3.1. Antisymmetric problem
It is expedient to choose in the displacement-stress representations (17)-(23) only one harmonic function
such
that:
φ1  x 1 , x 2 ,z 1   f  x 1 , x 2 ,z 1 
φ2  x 1 , x 2 ,z 2   f  x 1 , x 2 ,z 2 
(41)
φ3  x 1 , x 2 ,z 3   0
0
u3  u3   ε3  ω2 x 1  ω1x 2 ,  x 1 , x 2  S
(30)
Then the corresponding displacement and stress components
become:
u1  u2  0 ,  x 1 , x 2  R 2
(31)
uα   1β f  x 1 , x 2,zα , , α  1,2
β
σ 33  0,  x 1 , x 2  R 2  S
(32)
and are supplemented by the corresponding equilibrium
conditions to determine the unknown parameters
:






 σ 33 x 1 , x 2,0  σ 33 x 1, x 2,0  dx 1dx 2  0
S

 x 3α σ 33 x 1 , x 2,0
S




 σ 33 x 1 , x 2,0  dx 1dx 2  0

(33)
(34)
The symmetric part (B) of the problem leads to the following
mixed conditions
0
u1  u1   ε1  ω3 x 2 ,  x 1 , x 2  S
(35)
0
u2  u2   ε2  ω3 x 1 ,  x 1 , x 2  S
(36)
u3  0,  x 1 , x 2  R 2
(37)
2
σ 31  σ 32  0,  x 1 , x 2  R  S
(38)
and additional conditions to determine the unknown parameters
:
142
u3   1β mβ t β

f x 1 , x 2 ,z β
(42)

(43)
z β


2f x 1 , x 2,z β
σ 3α
β
  1 1  mβ t β
, α  1,2
c 44
z β x α


σ 33
β
  1 1  mβ
c 44




2f x 1 , x 2 ,z β

(45)
2
z β

σ 11  c 44  1β 1  mβ t β2
2f
z β2
(44)


 , 22
(46)
β
 c 11  c 12  1 f x 1 , x 2 ,z β 




σ 22  c 44  1β 1  mβ t β2
2f
z β2


 , 11
(47)
β
 c 11  c 12  1 f x 1 , x 2 ,z β 




σ 12  c 11  c 12   1β f x 1 , x 2,z β 

, 12
In the limit as
, the above equations reduce to:
(48)
acta mechanica et automatica, vol.7 no.3 (2013), DOI 10.2478/ama-2013-0024
u1  u2  0
(49)
u3   m2t 2  m1 t 1  f ,3  x 1 , x 2, x 3  x 3  0
(50)
σ 3α
  m2 t 2  m1 t 1 t 2 t 1  f ,3α  x 1 , x 2 , x 3  x  0
3
c 44
(51)

σ 33
  m2  m1  f ,33  x 1 , x 2 , x 3  x  0
3
c 44


(52)
σ 11  σ 22  c 44 m2 t 2 2  m1 t 1 2  t 2 2  t 1 2 


 f ,33  x 1 , x 2 , x 3  x  0
3
σ 12  0
 0  ε  ω x  ω x
3
2 1
1 2
m2 t 2  m1 t 1
(55)
(56)
This problem is substantially more complex (see the corresponding skew-symmetrical crack problem considered in Kassir
and Sih (1975)). We choose the potential functions in the general
representation (16)-(22) such that:
m2t 2
m2t 2
G1, 1  H 1, 2  
F1
m2t 2  m1t 1
m2t 2  m1t 1
m1t 1
m1t 1
φ2 
G2, 1  H 2, 2 
F2
m2t 2  m1t 1
m2t 2  m1t 1
φ3  G3 , 2  H 3 , 1  F3

φ1  



m2t 2  m1t 1
t 1t 2
u3 
m2t 2  m1t 1
σ 3α
c 44

 γα F3 ,γ , α  1,2
 F2 F1 



 z 2 z 1 

t 1t 2
F2
F 
 1  m2  1 
1  m1 
m2t 2  m1t 1 
z 2
z 1  , α
 γα t 3
σ 33
c 44 t 1t 2 
2F2
2F1 
 t 1 1  m2 
t 2 1  m1 

m2t 2  m1t 1 
z 22
z 12 
2F3
, α  1,2
z 3 x γ
1

c 44 m2t 2  m1t 1
(63)
  m1t 1 F2  m2t 2F1  , 11

 c 11  c 12  
 F3 , 12 
m
t

m
t
22
11


 m1t 1 F2  m2t 2F1  , 12
σ 12

 0,5 F3 , 11  F3 , 22
c 11  c 12
m2t 2  m1t 1

(64)
It is seen that the above expressions simplify across the plane
since the subscripts j can be dropped for
and ,
so we get
. Introducing
the notation:
g  x 1 , x 2, x 3  
G  x 1 , x 2 , x 3 
x 3
(65)
H  x 1 , x 2 , x 3 
x 3
the displacements and stresses in the plane of symmetry are
found as:

2F2
2F1 
t 2 1  m2 
t 1 1  m1 


z 22
z 12 
u2  h , 3  x 1 , x 2 , x 3  x 3  0
(57)
(58)
(66)
u3  0

σ 31
 C *  g , 33  κ g , 22  h , 12 

 x 3  0
c 44


where
and
,
are harmonic functions in the corresponding systems of coordinates. Then, the corresponding displacement-stress field becomes:
m1t 1 F2  m2t 2F1  , α
(62)
u1   g , 3  x 1 , x 2 , x 3   x 3  0
3.2. Symmetric problem
uα 
σ 22 
h  x 1 , x 2, x 3  
 for
f ,33  x 1 , x 2, x 3 

 x 3  0  0

2F2
2F1 
 t 1 1  m2 
t 2 1  m1 

z 22
z 12 


A glance at equations (37)-(39) reveals that the potential
is governed by:
 for
u3
c 44 t 1t 2
m2t 2  m1t 1
  m1t 1 F2  m2t 2F1  , 22

 c 11  c 12  
 F3 , 12 
m
t

m
t
22
11


(53)
(54)
f , 3  x 1 , x 2 , x 3  

 x 3  0 
σ 11 

(67)
σ 32
 C * h , 33  κ h , 11 g , 12 

 x 3  0
c 44
(68)
σ 33
  D *  g , 31  h , 32 
x3 0
c 44
(69)
σ 11  E *  g , 31  h , 32  x  0  c 11  c 12  h , 32 
x 3 0
3
(70)
σ 22  E *  g , 31  h , 32  x  0  c 11  c 12   g , 31 
x3  0
3
(71)
σ 12  0,5 c 11  c 12   g , 32  h , 31  x  0
3
(72)


where:
(59)
(60)
(61)
c 11c 33 t 
t t  m  m1 
C* 1 2 2

m2t 2  m1t 1
c 11c 33  c 44
t
κ 1 3
C*
D* 
(73)
A A c 
A
c
, E *  11   44
c 33 c 11c 33  c 44
c 11c 33  c 44
It follows immediately from the boundary conditions (35)-(38)
that the unknown potentials and can be determined from:
143
Andrzej Kaczyński
An Anticrack in Transversely Isotropic Space
 0

 g , 3  x 1 , x 2 , x 3 


 x 3  0   u1  ε1  ω3 x 2 
  x 1 , x 2  S (74)
0


h, 3  x 1 , x 2 , x 3 


u

ε

ω
x
2
3 1
2

 x 3  0





 g , 33  κ g , 22  h , 12 
 0

 x 3  0


2
  x 1 , x 2  R  S
h , 33  κ h , 11 g ,12 
 0

 x 3  0


(75)
where ξ  ξ 1 ,ξ 2  S , ξ  ξ 1 2  ξ 2 2 .
From the boundary condition (56), in view of (52), one finds
immediately that:
Af ξ 1 ,ξ 2  
1



4π 2c 44  m2  m1  S
σ 33 η1 ,η2  exp i ηα ξ α  dη1dη2 (77)

Accordingly, the conditions in equation (75) yield a system
for the remaining unknown functions
and
:
 κ ξ 2
κ ξ 1ξ 2 
2
1 
2
2

  Ag ξ 1 ,ξ 2  
ξ
ξ



2   A ξ ,ξ  

κ
ξ


κ
ξ
ξ
1  h 1 2 
1 2

1
2
2 

ξ
ξ



1
4π 2c 44 C *







S
(78)
which has the solution:
S
144
κ
dS ξ
2







S
(82)

σ 31  ξ  x 1  ξ 1  x 2  ξ 2 
x ξ
3
dS ξ
S
where
and
is a distance between the field point
and the
integration point
. Moreover, in deriving the above
expressions, use has been made of the following integrals
(Erdelyi, 1954):
exp i  x α  ηα ξ α 
dξ 1dξ 2 
ξ
R2
2π
2
 x 1  η1 
2
exp i  x α  ηα ξ α  ξ 1 
dξ 1dξ 2 
3
ξ


R2








2
exp i  x α  ηα ξ α  ξ 2 
dξ 1dξ 2 
3
ξ


R2








2
  x 2  η2 
2π  x 2  η2 
3
 x 1  η1 2   x 2  η2 2 

2π  x 1  η1 
3
 x 1  η1 2   x 2  η2 2 

exp i  x α  ηα ξ α  ξ 1ξ 2
 2π  x 1  η1  x 2  η2 
dξ 1dξ 2 
3
3
ξ

2
2
  x 1  η1    x 2  η2  
R2








Now enforcing the displacement boundary conditions (55)
and (74), we arrive at the governing singular integral equations
of Newtonian potential type to determine the interface stresses
:
B
(79)








σ 33 ξ 1 ,ξ 2 dξ 1dξ 2
2
x1 ξ 1 








S





2
 x2 ξ 2 

(80)
(85)
 0  ε  ω x  ω x
3
2 1
1 2
 u3



 x 2  ξ 2 2
1  κ

 x 1  ξ 1 2   x 2  ξ 2 2 
 x 1  ξ 1 2   x 2  ξ 2 2 
σ 31 ξ 1 ,ξ 2 

σ ξ ,ξ  x  ξ  x 2  ξ 2 
κ 32 1 2 1 1
3

2
2
 x1 ξ 1   x2 ξ 2  
Equations (77) and (79) may be substituted into (76) to give:

3


S
S
S

x ξ

 σ
ξ 
 σ
32  ξ  x 1  ξ 1 
2π t 3c 44 h , 3  x    32 dS ξ  κ 
dS ξ


3
 x  ξ

x ξ

A
 ξ  dS
 σ
2π c 44  m2  m1  f , 3  x     33
ξ

 x  ξ

(81)

σ 32  ξ  x 1  ξ 1  x 2  ξ 2 
(83)
σ  η ,η  
 31 1 2  exp  i ηα ξ α   dη1dη2


σ  η ,η  
 32 1 2 
 κ ξ 2
κξ ξ 
1
1 
 1 2 
2
2

 Ag  ξ   
ξ
ξ
2


4π c 44 C * 1  κ  

κ ξ 2 2 
 Ah  ξ    κ ξ 1ξ 2

1
2
2 

ξ
ξ






 σ
η ,η 
 31 1 2  exp  i ηα ξ α   dη1dη2
 





 
 σ 32 η1 ,η2 

 
S






S






S
2


(76)


S
κ
It is easy to seen from a comparison of the above conditions
that the formulation given by Eqs. (74) and (75) is inverse to that
for the crack problem involving shear tractions (Kassir and Sih,
1975).
To obtain integral equations, the two-dimensional Fourier
technique will be used (see the method developed by Kaczyński
(1999)). The space harmonic functions
and
are
represented by Fourier’s integrals (Sneddon, 1972):

 A ξ  
 f  x   
exp  x 3 ξ  i  x α ξ α   f


 

 Ag  ξ   dS ξ
 g  x    
2

ξ


 h  x   

 
 Ah  ξ  



 σ
ξ 
 σ
31  ξ  x 2  ξ 2 
2π t 3c 44 g , 3  x    31 dS ξ  κ 
dS ξ


3
 x  ξ

x ξ





 0
dξ 1dξ 2  u1  ε 1  ω3 x 2



(84)
acta mechanica et automatica, vol.7 no.3 (2013), DOI 10.2478/ama-2013-0024
B








S





Where:



 x 1  ξ 1 2
1  κ

2
2
 x 1  ξ 1    x 2  ξ 2  
 x 1  ξ 1 2   x 2  ξ 2 2 
σ 32 ξ 1 ,ξ 2 
D



σ 31 ξ 1 ,ξ 2  x 1  ξ 1  x 2  ξ 2  
 0
κ
dξ dξ  u2  ε 2  ω3 x 1
3  1 2

2
2 
 x1 ξ 1   x2 ξ 2   

 
A
B
1
2π c 44 t 3

1
(87)
π 2c 44 c 11  c 12 
It is worth mentioning that the results obtained can be used
in the case of isotropic material (see the relations (16)). Then the
constants in the governing equations become:
κ
λ μ
1

2  λ  2μ  4 1 ν 
A
λ  3μ
3  4ν

4π μ  λ  2μ  8π μ 1 ν 
B
1
2πμ
(88)
which are in agreement with those obtained by Silovanyuk (1984).
After the stresses
acting on the inclusion side
have
been determined from the solution of the above integral
equations, the full-space elastic field can be calculated from the
harmonic potentials
via relations (80)-(83) with the use of
Eqs. (42)-(48) and (57)-(64). For completeness, the kinematical
parameters of the inclusion
and
can be found from the
conditions (33), (39) and (34), (40), respectively.
For an arbitrary simply connected domain
bounded
by a smooth contour the governing equations can be solved
by an analytic-numerical method developed in Kit et al. (1989).
Explicit solutions are possible for elliptical shapes of
by assuming polynomial right-hand sides. The typical case
of a circular anticrack in a uniform normal tension at infinity will be
considered in the next section.
4. EXAMPLE
The stress state of a transversely isotropic space containing
in the plane of isotropy a sealed circular rigid inclusion (anticrack)
{
} is investigated
√
by assuming that:
    σ  const.  0, σ     σ     0
σ 33
  3
31  
32  
t
t
t
(89)
The displacement solution to the basic equations (7) and (8)
of 0-problem with conditions (89) is readily obtained as:
0
uα   x 1 , x 2 , x 3   D σ 3 x α , α  1, 2
 0 x , x , x  D σ x
 1 2 3 3 3 3
u3
c 11  c 12
(91)
2
c 33 c 11  c 12   2c 13 
Now invoking the displacement boundary conditions (see Eqs.
(27)-(29)), we are dealing with the symmetric part of the perturbed
problem described by Eqs. (35)-(40) in which
is given by Eqs. (73), and the material constants ,
c c c
m2 t 2  m1t 1
 11 33 44
2π c 44  m2  m1 
2π c 44c 33 t 
2
c 33 c 11  c 12   2 c 13 
D3 
(86)
in which
are:
c 13
uα  x 1, x 2   D σ3 x α , α  1, 2
The exact analytical solution of the governing system (85) and
(86) with the RHS given by (92) is achieved if we take the unknown stresses in the form
b b x b x

σ 3α  x 1 , x 2   0α 1α 1 2α 2 , α  1, 2
a2  r 2
(93)
where bjα , j  0, 1, 2, α  1, 2 are unknown constants.
Substituting these expressions into Eqs. (85) and (86),
and next using the formulas for resulting integrals given by Vorovich et al. (1974), the equality of two polynomials of the first
order is obtained. Equating the terms of the left and right hands
with the same powers of
and
yields a system of linear
algebraic equations, solving which we get the following formulas
for the unknown coefficients
:
b0α  
4c 44 t 3
ε , α  1, 2
π 2  κ  α
4
b12  b21  c 44 t 3 ω3
π
(94)
4
b11  b22   c 44 DC * σ 3
π
Now the equilibrium conditions provided in (39) and (40) can
be employed, from which we get (as might be expected):
(95)
ε1  ε2  ω3  0
Thus, expression (93) can be simplified as:

σ 3α  x 1 , x 2   
4c 44DC * σ 3 x α
, α  1, 2
π a2  r 2
(96)
From above it follows that the problem in hand is axially symmetric.
On substitution of formulas (96) into Eqs. (81) and (82), we
find:
g, 3 x 
h, 3  x  
2DC * σ 3
π 2t 3
2 DC * σ 3
π 2t 3
Here
ψα  x  
ψ1  κ ψ1, 2 x 2  κ ψ2, 2 x 1 

ψ2  κ ψ2, 2 x 1  κ ψ1, 1x 2

(97)
are the simple-layer potentials defined by:






S
(90)
(92)
ξ α dξ 1 dξ 2
2
2
x  ξ a 2  ξ 1   ξ 2 
(98)
They can be expressed in elementary functions by utilizing
the results of Fabrikant (1989, 1991) as:
145
Andrzej Kaczyński
An Anticrack in Transversely Isotropic Space

a  a
ψα  x   π x α  sin1   
 l2 

5. CONCLUSIONS
 l 2 2  a 2 
 l 2 2 
(99)
where l 2  0,5   r  a 2   x 3 2   r  a 2   x 3 2  .


Having the above explicit expressions, a complete solution
to the problem under study is available. Since the resulting formulas are lengthy, we omit them here to save the space of the paper.
To investigate the singular behavior of anticrack-border stresses,
however, the shear radial and normal stresses in the plane
are calculated as follows:
 

σ 3r r ,0
βr σ3 r

, 0r a

  π a2  r 2

r a
 0,
(100)
 1
0r a
 2 β3σ 3 ,

σ 33 r ,0  

a
 a 
 β3σ 3 
 sin1    , r  a
 π  2 2
 r  
 r a

 
in which
βr 
β3 
(101)
are given by:
4c 13 c 11c 44 c 11c 33  c 13 2c 44  c 13   2c 44 c 11c 33 
c c  c   2c 2 
13 
 33 11 12

4c 13 c 44


c 11c 33  c 44
c 11c 33  c 13
c c  c   2c 2 
13 
 33 11 12



c 11c 33  c 44


(102)
(103)
These results reveal that the stresses near the anticrack front
have the classical singularity
as in the fracture
mechanics of conventional elastic materials. Strictly speaking,
singularities in
occur at the points on the edge of the disc
where
, and in
 at the points exterior to the disc
where
. It indicates that there are two major mechanisms
controlling the material cracking around the inclusion front:
‒ exfoliation of the material from the surface of the inclusion
described by the stress singularity coefficients:

S II  lim
r  a
 
2π a  r  σ 3r r ,0  β r σ 3 a / π
(104)
‒ mode I fracture in the immediate vicinity of the edge of the
disc characterized by the stress intensity factor:
K I  lim
r  a
2π  r  a  σ 33  r ,0  β3σ 3 a / π
(105)
The above-mentioned parameters can be used in conjunction
with a suitable failure criterion.
Finally, in the special case of isotropy (see (16)), we arrive
at the solution in which the constants
become:
βr 
4λ  λ  2μ 
8ν 1 ν 

λ

3
μ
3
λ

2
μ
1


  ν 3  4ν 
(106)
β3 
4ν 1  2ν 
4λ μ

 λ  3μ 3λ  2μ  1 ν 3  4ν 
(107)
in complete agreement with Kassir and Sih (1968).
146
In the present paper we have studied the elastostatic threedimensional problem of an anticrack of arbitrary shape embedded
in a transversely isotropic space and subjected to external loads.
Using the method of potential functions, the mixed boundary-value
problems in the antisymmetric and symmetric statement have
been reduced to some mixed problems of potential theory. Further, the governing boundary integral equations were obtained
with the unknown stress jumps across the rigid inclusion.
As an illustration, a closed-form solution was given and discussed
for a circular rigid inclusion subjected to normal tension at infinity.
The analytical expressions of stress fields in the anticrack plane
show the characteristic
singular behaviour near the edge
of the disc and indicate that either the matrix fractures near the
inclusion according to mode I or the mechanism of shear fracture
acts (separation of the material from the inclusion surface).
REFERENCES
1. Berezhnitskii L.T., Panasyuk V.V. , Stashchuk N.G. (1983), The
Interaction of Rigid Linear Inclusions and Cracks in a Deformable
Body (in Russian), Naukova Dumka, Kiev.
2. Chaudhuri R.A. (2003), Three-dimensional asymptotic stress field
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